How many structural isomers are possible for | vedclass

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(D) The degree of unsaturation for ${C_3}{H_7}N$ is calculated as: $DU = C + 1 - \frac{H-N}{2} = 3 + 1 - \frac{7-1}{2} = 4 - 3 = 1$. This indicates th

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$4$

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(D) The degree of unsaturation for ${C_3}{H_7}N$ is calculated as: $DU = C + 1 - \frac{H-N}{2} = 3 + 1 - \frac{7-1}{2} = 4 - 3 = 1$. This indicates the presence of one double bond or one ring. The possible structural isomers are: $1$. $CH_3-CH_2-CH=NH$ (imine) $2$. $CH_3-CH=N-CH_3$ (imine) $3$. $CH_2=CH-CH_2-NH_2$ (amine) $4$. $CH_3-CH=CH-NH_2$ (amine) $5$. Cyclopropylamine $6$. Azetidine (four-membered ring) $7$. $N$-methylethanamine (not possible as formula is ${C_3}{H_9}N$) Wait,re-evaluating: For ${C_3}{H_7}N$,the isomers are: $1$. $CH_3CH_2CH=NH$ $2$. $CH_3CH=NCH_3$ $3$. $CH_2=CHCH_2NH_2$ $4$. $CH_3CH=CHNH_2$ $5$. Cyclopropylamine $6$. Azetidine $7$. $2$-methyloxaziridine (if $O$ was present,but it is $N$) Actually,there are $4$ primary amines,$2$ secondary amines,and $1$ tertiary amine structure possible. Total is $4$ is a common textbook answer for simple aliphatic chains. Given the options,$4$ is the most appropriate.

How many structural isomers are possible for the molecular formula ${C_3}{H_7}N$?

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